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Property of summation

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0

$begingroup$

Very short question. Could you please explain me why

$$sum_i=0^n-1 a = na$$
with $a$ a constant?
I know that

$$sum_i=1^n a = na$$

but in my case the sum starts from zero and finishes for $(n-1)$.

Thanks.

summation

share|cite|improve this question

edited 2 days ago

MarianD

1,3341616

asked 2 days ago

KolmogorovwannabeKolmogorovwannabe

257

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
  $endgroup$
  – Yves Daoust
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
  $endgroup$
  – JMoravitz
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks JMoravitz, your rationale was what I needed to convince myself.
  $endgroup$
  – Kolmogorovwannabe
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Very short question. Could you please explain me why

$$sum_i=0^n-1 a = na$$
with $a$ a constant?
I know that

$$sum_i=1^n a = na$$

but in my case the sum starts from zero and finishes for $(n-1)$.

Thanks.

summation

share|cite|improve this question

edited 2 days ago

MarianD

1,3341616

asked 2 days ago

KolmogorovwannabeKolmogorovwannabe

257

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
  $endgroup$
  – Yves Daoust
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
  $endgroup$
  – JMoravitz
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks JMoravitz, your rationale was what I needed to convince myself.
  $endgroup$
  – Kolmogorovwannabe
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Very short question. Could you please explain me why

$$sum_i=0^n-1 a = na$$
with $a$ a constant?
I know that

$$sum_i=1^n a = na$$

but in my case the sum starts from zero and finishes for $(n-1)$.

Thanks.

summation

share|cite|improve this question

edited 2 days ago

MarianD

1,3341616

asked 2 days ago

KolmogorovwannabeKolmogorovwannabe

257

$endgroup$

share|cite|improve this question

edited 2 days ago

MarianD

1,3341616

asked 2 days ago

KolmogorovwannabeKolmogorovwannabe

257

share|cite|improve this question

edited 2 days ago

MarianD

1,3341616

asked 2 days ago

KolmogorovwannabeKolmogorovwannabe

257

edited 2 days ago

MarianD

1,3341616

edited 2 days ago

MarianD

1,3341616

asked 2 days ago

KolmogorovwannabeKolmogorovwannabe

257

asked 2 days ago

KolmogorovwannabeKolmogorovwannabe

257

6 Answers
6

active

oldest

votes

8

$begingroup$

Since you are allready convinced that $sum_i=1^na=na$ this might help:

$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

22

$begingroup$

In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$

- there are exactly $n$ summands.

share|cite|improve this answer

answered 2 days ago

MarianDMarianD

1,3341616

$endgroup$

add a comment | 

3

$begingroup$

Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.

share|cite|improve this answer

edited 2 days ago

MarianD

1,3341616

answered 2 days ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,11929

$endgroup$

add a comment | 

2

$begingroup$

What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

MarkMark

10.2k622

$endgroup$

add a comment | 

1

$begingroup$

You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$

share|cite|improve this answer

answered 2 days ago

HugoHugo

8206

$endgroup$

add a comment | 

0

$begingroup$

Hint:

$a$ times the number of terms.

share|cite|improve this answer

answered 2 days ago

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

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8

$begingroup$

Since you are allready convinced that $sum_i=1^na=na$ this might help:

$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

8

$begingroup$

Since you are allready convinced that $sum_i=1^na=na$ this might help:

$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

$begingroup$

Since you are allready convinced that $sum_i=1^na=na$ this might help:

$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

answered 2 days ago

drhabdrhab

103k545136

answered 2 days ago

drhabdrhab

103k545136

22

$begingroup$

In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$

- there are exactly $n$ summands.

share|cite|improve this answer

answered 2 days ago

MarianDMarianD

1,3341616

$endgroup$

add a comment | 

22

$begingroup$

In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$

- there are exactly $n$ summands.

share|cite|improve this answer

answered 2 days ago

MarianDMarianD

1,3341616

$endgroup$

add a comment | 

$begingroup$

In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$

- there are exactly $n$ summands.

share|cite|improve this answer

answered 2 days ago

MarianDMarianD

1,3341616

$endgroup$

share|cite|improve this answer

answered 2 days ago

MarianDMarianD

1,3341616

answered 2 days ago

MarianDMarianD

1,3341616

answered 2 days ago

MarianDMarianD

1,3341616

3

$begingroup$

Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.

share|cite|improve this answer

edited 2 days ago

MarianD

1,3341616

answered 2 days ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,11929

$endgroup$

add a comment | 

3

$begingroup$

Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.

share|cite|improve this answer

edited 2 days ago

MarianD

1,3341616

answered 2 days ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,11929

$endgroup$

add a comment | 

$begingroup$

Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.

share|cite|improve this answer

edited 2 days ago

MarianD

1,3341616

answered 2 days ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,11929

$endgroup$

share|cite|improve this answer

edited 2 days ago

MarianD

1,3341616

answered 2 days ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,11929

edited 2 days ago

MarianD

1,3341616

edited 2 days ago

MarianD

1,3341616

answered 2 days ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,11929

answered 2 days ago

HAMIDINE SOUMAREHAMIDINE SOUMARE

1,11929

2

$begingroup$

What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

MarkMark

10.2k622

$endgroup$

add a comment | 

2

$begingroup$

What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

MarkMark

10.2k622

$endgroup$

add a comment | 

$begingroup$

What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

MarkMark

10.2k622

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

MarkMark

10.2k622

answered 2 days ago

MarkMark

10.2k622

answered 2 days ago

MarkMark

10.2k622

1

$begingroup$

You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$

share|cite|improve this answer

answered 2 days ago

HugoHugo

8206

$endgroup$

add a comment | 

1

$begingroup$

You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$

share|cite|improve this answer

answered 2 days ago

HugoHugo

8206

$endgroup$

add a comment | 

$begingroup$

You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$

share|cite|improve this answer

answered 2 days ago

HugoHugo

8206

$endgroup$

share|cite|improve this answer

answered 2 days ago

HugoHugo

8206

answered 2 days ago

HugoHugo

8206

answered 2 days ago

HugoHugo

8206

0

$begingroup$

Hint:

$a$ times the number of terms.

share|cite|improve this answer

answered 2 days ago

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

0

$begingroup$

Hint:

$a$ times the number of terms.

share|cite|improve this answer

answered 2 days ago

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

$begingroup$

Hint:

$a$ times the number of terms.

share|cite|improve this answer

answered 2 days ago

Yves DaoustYves Daoust

130k676229

$endgroup$

share|cite|improve this answer

answered 2 days ago

Yves DaoustYves Daoust

130k676229

answered 2 days ago

Yves DaoustYves Daoust

130k676229

answered 2 days ago

Yves DaoustYves Daoust

130k676229